Table of Contents
Chapter 6

— David Wagner 2007/06/29 15:40

Problem 1.

{s^2 h}/{27 D} = {10^2 * 21.9}/{27*24}=3.4

3.4 BCY/ft

Problem 2.

T={4000/{1.5*5280}}+0.30=0.35 min;

Productivity P={2.22DWLE}/T={2.22*2*37*400*0.70}/3.33 =1120

1120 BCY/hr

Unit Cost: $30/1120 = 0.12

$0.12/BCY

Problem 3.

[tbl 6-5] I=5*1.5=7.5A; [tbl 6-6] R_c={NR}/P={6*2.7}/5=2.76Ω; [tbl 6-7] R_{bus}={42/1000}*4.02=0.16Ω; R_{leads}=1200*2*{1.59/1000}=3.82Ω; R=2.76+0.16+3.81=6.74Ω;

E=IR=7.5*6.74=50.6

7.5 A at 50.6 V

Problem 4.

Increasing air pressure boosts productivity at the cost of increased wear and tear on the equipment. At all times the equipment must stay within safe operating limits.

Problem 5.

Graph for this Problem V_1={20-0}/{0.02-0}=1000 ft/sec; V_2={60-20}/{0.033-0.002}=3000 ft/sec; V_3={80-60}/{0.0367-0.0332}=5882 ft/sec; H_1={D_1/2}sqrt{{V_2 - V_1}/{V_2 + V_1}} = {20/2}sqrt{{3000 - 1000}/{3000 + 1000}}=7.1 ft

7.1 ft

Problem 6.

{8*8*18}/27=42.67 BCY/hole; 42.67/20=2.13 BCY/drill-ft; P=2.13*70=149.3 BCY/hr-drill;

Drilling Unit Cost: {40+70}/149.3=$0.74/BCY

Blasting Unit Cost: {42/149.3}+{4/42.6}+0.40*1=$0.78/BCY

Total Cost: $1.52/BCY

Problem 7.

Marginal is not rippable. [fig. 6-17]

Problem 8.

[tbl 6-5] I=6*3=18A; [tbl 6-6] R_c={NR}/P={8*2.4}/6=3.26Ω; [tbl 6-7] R_{bus}={52/1000}*4.02=0.21Ω; R_{leads}=1000*2*{2.53/1000}=5.06Ω; ΣR=8.53Ω;

E=IR=18*8.53=153.5V

18 A at 153.5 V

Discussion


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